The given diagram shows two semi-infinite lin | vedclass

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(C) For a semi-infinite line of charge with linear charge density $\lambda$,the electric field at a distance $r$ from the end of the line is given by

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$\hat j$

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(C) For a semi-infinite line of charge with linear charge density $\lambda$,the electric field at a distance $r$ from the end of the line is given by $E = \frac{k\lambda}{r}$ at an angle of $45^\circ$ to the line. However,using the symmetry of the given configuration,let the positive line of charge make an angle $	heta$ with the $x$-axis and the negative line of charge make an angle $	heta$ with the $x$-axis. The electric field due to the positive line of charge $(E_+)$ points away from the line,and the electric field due to the negative line of charge $(E_-)$ points towards the line. By symmetry,the components of the electric fields along the $x$-axis cancel each other out,while the components along the $y$-axis add up. Thus,the resultant electric field $E_R$ is directed along the positive $y$-axis. The unit vector along the positive $y$-axis is $\hat j$. Therefore,the correct option is $C$.

The given diagram shows two semi-infinite lines of charge having equal (in magnitude) linear charge density but with opposite signs. The electric field at any point on the $x$-axis for $(x > 0)$ is along the unit vector:

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